Reimann integral question

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July 8th 2011, 10:35 AM
CountingPenguins

Reimann integral question

Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

Why isn't g(x)= 0 why does it have to be f(x)? Why not both?
July 8th 2011, 10:53 AM
girdav

Re: Reimann integral question

What we suppose works for all Riemann-integrable function $g$ , and we only know that $\int_a^b fg =0$ .
July 8th 2011, 11:09 AM
Plato

Re: Reimann integral question

Quote:

Originally Posted by CountingPenguins

Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

From the given do you know that $\int_a^b {f^2 } = 0~?$

Is that possible if $f(t)\not=0$ for some $t\in[a,b]~?$
July 8th 2011, 12:29 PM
CountingPenguins

Re: Reimann integral question

f is continuous and g is integrable and the product of the integral of fg over the interval [a,b] is 0. So f(x) must be 0 for all x in [a,b], is all the information I have been given. I will start with the definition of continuous for f and integrable for g and then hope I see something useful after that.
July 8th 2011, 12:45 PM
Plato

Re: Reimann integral question

Did you see reply #3. That answers the question.
July 10th 2011, 01:49 PM
Also sprach Zarathustra

Re: Reimann integral question

Quote:

Originally Posted by CountingPenguins

Let f be continuous on [a,b] and suppose that, for every integrable function g defined on [a,b],∫_{a}^{b}fg=0. Prove that f(x)=0 for all x∈[a,b].

Why isn't g(x)= 0 why does it have to be f(x)? Why not both?

I came up with the following solution, I hope it is not so wrong. (Itwasntme)

Suppose that $f$ is not $0$ for all $x\in[a,b]$ , then there exists $x_0\in[a,b]$ such that $f(x_0)>0$ (If the is no such $x_0$ , there must be $x_1\in[a,b]$ such that $f(x_1)<0$ .In that case we will look at $-f$ ).

f is continuous function therefor exists $\delta>0$ for which $f(x)>\epsilon>0$ for all $x\in (x_0-\delta,x_0+\delta)$ .

Now we define $h(x)$ -'staircase'-characteristic function of open interval $(x_0-\delta,x_0+\delta)$ .

$h(x)=\left\{\begin{matrix}1 & x\in (x_0-\delta,x_0+\delta) \\ 0 & \text{otherwise} \end{matrix}\righ$

Hence, we will get:

$0=\int_{a}^{b}f(x)h(x)dx=\int_{a}^{x_0-\delta}f(x)h(x)dx+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+\int_{x_0+\delta}^{ b}f(x)h(x)dx=0+\int_{x_0-\delta}^{x_0+\delta}f(x)h(x)dx+0>2\epsilon\delta$ .

So, we got a contradiction to the fact that if $f$ is continuous on $[a,b]$ and h is staircase function $\int_{a}^{b}f(x)h(x)dx=0 \Rightarrow f\equiv 0$ .

But we also know that if $g$ is integrable function on $[a,b]$ , so for every $\epsilon_0>0$ exists staircase function $h$ so that

$\int_{a}^{b}|g(x)-h(x)|<\epsilon_0$

Combining these two results we will conclude the needed.
July 10th 2011, 03:35 PM
Plato

Re: Reimann integral question

Quote:

Originally Posted by Also sprach Zarathustra

I came up with the following solution, I hope it is not so wrong.

Your solution is correct.
Note that $f \text{ is continuous on }[a,b]$ so as such that means that both $f~\&~f^2$ is are integrable on $[a,b]$ .

By the given $\int_a^b {f \cdot f} = \int_a^b {f^2 } = 0$ .
But by your reasoning that must imply that $f \equiv 0$ .